Inequality 32(Vasile Cirtoaje)

The following inequality and its solution is dedicated to my beloved teacher, Christos Patilas…

Problem:

If \displaystyle a,b,c,d are positive real numbers, with \displaystyle abcd=1, prove that

\displaystyle \frac{1}{(1+a)(1+a^2)}+\frac{1}{(1+b)(1+b^2)}+\frac{1}{(1+c)(1+c^2)}+\frac{1}{(1+d)(1+d^2)}\geq 1.

Solution:

Without loss of generality assume that \displaystyle a\geq b\geq c\geq d. Then from Chebyshev’s inequality we have

\displaystyle \frac{1}{(1+a)(1+a^2)}+\frac{1}{(1+b)(1+b^2)}+\frac{1}{(1+c)(1+c^2)}\geq \frac{1}{3}\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\right).

Lemma (Vasile Cirtoaje):

If \displaystyle a\geq b\geq c\geq d and \displaystyle abcd=1 then it holds that

\displaystyle \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\geq \frac{3}{1+\sqrt[3]{abc}}.

Proof of the Lemma:

We know that \displaystyle \frac{1}{1+a}+\frac{1}{1+b}\geq \frac{2}{1+\sqrt{ab}}. So, it suffices to prove that \displaystyle \frac{1}{1+c}+\frac{2}{1+\sqrt{ab}}\geq \frac{3}{1+\sqrt[3]{abc}}.

Let us denote by \displaystyle x=\sqrt{ab},y=\sqrt[3]{abc}\Longrightarrow c=\frac{y^3}{x^2}. Substituting them to the above inequality we get that

\displaystyle \frac{1}{1+c}+\frac{2}{1+\sqrt{ab}}-\frac{3}{1+\sqrt[3]{abc}}=\frac{x^2}{x^2+y^3}+\frac{2}{1+x}-\frac{3}{1+y},

which reduces to the inequality

\displaystyle \frac{(x-y)^2\left[2y^2-y+x(y-2)\right]}{(1+x)(1+y)(x^2+y^3)},

which is obvious since \displaystyle 2y^2-y+(y-2)x\geq 2y^2-y+(y-2)y^3=y(y-1)(y^2-y+1)\geq 0.

Back to our inequality now, from the above lemma we deduce that:

\displaystyle \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\geq \frac{3}{1+\sqrt[3]{abc}}\wedge \frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\geq \frac{3}{1+\sqrt[3]{a^2b^2c^2}}.

For convenience denote by \displaystyle k the \displaystyle \sqrt[3]{abc}. Therefore we have that

\displaystyle \frac{1}{(1+a)(1+a^2)}+\frac{1}{(1+b)(1+b^2)}+\frac{1}{(1+c)(1+c^2)}\geq \frac{3}{(1+k)(1+k^2)}.

Thus it remains to prove that

\displaystyle \frac{3}{(1+k)(1+k^2)}+\frac{1}{(1+d)(1+d^2)}\geq 1.

But \displaystyle abcd=1. So, the last fraction is of the form \displaystyle \frac{1}{\left(1+\frac{1}{k^3}\right)\left(1+\frac{1}{k^6}\right)}.

After that we get

\displaystyle \frac{1}{\left(1+\frac{1}{k^3}\right)\left(1+\frac{1}{k^6}\right)}+\frac{3}{(1+k)(1+k^2)}\geq 1.

Conclusion follows from the obvious inequality  \displaystyle \frac{(k-1)^2(2k^4+k^3+k+2)}{(k^3+1)(k^6+1)}\geq 0

Inequality 30(APMO 2005)

Problem:

Let \displaystyle x,y,z be positive real numbers such that \displaystyle xyz=8. Prove that

\displaystyle \frac{x^2}{\sqrt{(x^3+1)(y^3+1)}}+\frac{y^2}{\sqrt{(y^3+1)(z^3+1)}}+\frac{z^2}{\sqrt{(z^3+1)(x^3+1)}}\geq\frac{4}{3}.

Solution:

From the AM-GM ineuqality we know that

\displaystyle \frac{1}{\sqrt{x^3+1}}=\frac{1}{\sqrt{(x+1)(x^2-x+1)}}\geq \frac{2}{(x+1)+(x^2-x+1)}=\frac{2}{x^2+2}.

Doing that cyclic over the \displaystyle 3 fractions we get that

\displaystyle \sum_{cyc}\frac{x^2}{\sqrt{(x^3+1)(y^3+1)}}\geq \frac{4x^2}{(x^2+2)(y^2+2)}.

So, it suffices to prove that

\displaystyle \frac{4x^2}{(x^2+2)(y^2+2)}\geq \frac{4}{3}, or \displaystyle 3\sum_{cyc}x^2(z^2+2)\geq \prod_{cyc}(x^2+2).

After expanding, the inequality is equivalent to

\displaystyle 2\sum_{cyc}x^2+\sum_{cyc}x^2y^2\geq 72.

But the last relation holds due to the AM-GM inequality and from the hypothesis, since \displaystyle \sum_{cyc}x^2y^2\geq 3\sqrt[3]{8^4}=48\wedge 2\sum_{cyc}x^2\geq 6\sqrt[3]{8^2}=24.

Adding up these \displaystyle 2 relations we get the desired result, Q.E.D.

Inequality 27(Vo Quoc Ba Can)

Problem:

Let \displaystyle a,b,c be positive real numbers. Prove that

\displaystyle \frac{a^{b+c}}{(b+c)^2}+\frac{b^{c+a}}{(c+a)^2}+\frac{c^{a+b}}{(a+b)^2}\geq \frac{3}{4}

Solution:

It holds for every positive number \displaystyle x that

\displaystyle \frac{1}{x^2}\geq \frac{1}{x^{2x}}.

Back to our inequality now, rewrite it as follows:

\displaystyle \frac{a^{b+c}}{(b+c)^2}+\frac{b^{c+a}}{(c+a)^2}+\frac{c^{a+b}}{(a+b)^2}\geq \frac{3}{4}\Longrightarrow \sum_{cyc}\frac{4a^{b+c}}{(b+c)^2}\geq 3\Longrightarrow \sum_{cyc}\frac{a^{b+c}}{\left(\frac{b+c}{2}\right)^{2}}\geq 3.

Using the above result we have that

\displaystyle \sum_{cyc}\frac{a^{b+c}}{\left(\frac{b+c}{2}\right)^{2}}\geq \sum_{cyc}\left(\frac{a}{\frac{b+c}{2}}\right)^{b+c}=\sum_{cyc}\left(\frac{2a}{b+c}\right)^{b+c}.

Let us now make the final step for the proof of our inequality. From Bernulli’s inequality we get that

\displaystyle \sum_{cyc}\left(\frac{2a}{b+c}\right)^{b+c}=\sum_{cyc}\left[\left(\frac{2a}{b+c}-1\right)+1\right]^{b+c}\geq \sum_{cyc}(1+2a-b-c)=3, Q.E.D.